Question

已知数列{An} A1=1 ,A2=4,An=A(n-2)+2,求An和Sn

Answer 1

a(2n+1)=a(2n-1)+2
{a(2n-1)} 是首项为a(1)=1,公差为2的等差数列.
a(2n-1)=1+2(n-1)=2n-1.
a(2n+2)=a(2n)+2
{a(2n)}是首项为a(2)=4,公差为2的等差数列.
a(2n)=4+2(n-1)=2n+2.
综合,有
a(2n-1)=2n-1,
a(2n)=2n+2.
s(2n)=[a(1)+a(2)]+[a(3)+a(4)]+...+[a(2n-1)+a(2n)]
=[1+4]+[3+6]+...+[2n-1+2n+2]
=5+9+...+(4n+1)
=4n(n+1)/2 + n
=2n(n+1)+n
s(2n-1)=s(2n)-a(2n)=2n(n+1)+n-(2n+2)=2(n+1)(n-1)+n
综合,有
s(2n-1)=2(n+1)(n-1)+n
s(2n)=2n(n+1)+n